Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{4x - 28}{x + 3} \div \dfrac{x^2 - 14x + 49}{-4x - 12} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{4x - 28}{x + 3} \times \dfrac{-4x - 12}{x^2 - 14x + 49} $ First factor the quadratic. $q = \dfrac{4x - 28}{x + 3} \times \dfrac{-4x - 12}{(x - 7)(x - 7)} $ Then factor out any other terms. $q = \dfrac{4(x - 7)}{x + 3} \times \dfrac{-4(x + 3)}{(x - 7)(x - 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ 4(x - 7) \times -4(x + 3) } { (x + 3) \times (x - 7)(x - 7) } $ $q = \dfrac{ -16(x - 7)(x + 3)}{ (x + 3)(x - 7)(x - 7)} $ Notice that $(x + 3)$ and $(x - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -16\cancel{(x - 7)}(x + 3)}{ (x + 3)\cancel{(x - 7)}(x - 7)} $ We are dividing by $x - 7$ , so $x - 7 \neq 0$ Therefore, $x \neq 7$ $q = \dfrac{ -16\cancel{(x - 7)}\cancel{(x + 3)}}{ \cancel{(x + 3)}\cancel{(x - 7)}(x - 7)} $ We are dividing by $x + 3$ , so $x + 3 \neq 0$ Therefore, $x \neq -3$ $q = \dfrac{-16}{x - 7} ; \space x \neq 7 ; \space x \neq -3 $